Lemma 104.3.2. With assumptions and notation as in Cohomology of Stacks, Lemma 103.15.1. We have

$g^{-1} \circ Rf_* = Rf'_* \circ (g')^{-1} \quad \text{and}\quad L(g')_! \circ (f')^{-1} = f^{-1} \circ Lg_!$

on unbounded derived categories (both for the case of modules and for the case of abelian sheaves).

Proof. Let $\tau = {\acute{e}tale}$ (resp. $\tau = fppf$). Let $\mathcal{F}$ be an abelian sheaf on $\mathcal{X}_\tau$. By Cohomology of Stacks, Lemma 103.15.3 the canonical (base change) map

$g^{-1}Rf_*\mathcal{F} \longrightarrow Rf'_*(g')^{-1}\mathcal{F}$

is an isomorphism. The rest of the proof is formal. Since cohomology of abelian groups and sheaves of modules agree we also conclude that $g^{-1} Rf_*\mathcal{F} = Rf'_* (g')^{-1}\mathcal{F}$ when $\mathcal{F}$ is a sheaf of modules on $\mathcal{X}_\tau$.

Next we show that for $\mathcal{G}$ (either sheaf of modules or abelian groups) on $\mathcal{Y}_{lisse,{\acute{e}tale}}$ (resp. $\mathcal{Y}_{flat,fppf}$) the canonical map

$L(g')_!(f')^{-1}\mathcal{G} \to f^{-1}Lg_!\mathcal{G}$

is an isomorphism. To see this it is enough to prove for any injective sheaf $\mathcal{I}$ on $\mathcal{X}_\tau$ the induced map

$\mathop{\mathrm{Hom}}\nolimits (L(g')_!(f')^{-1}\mathcal{G}, \mathcal{I}[n]) \leftarrow \mathop{\mathrm{Hom}}\nolimits (f^{-1}Lg_!\mathcal{G}, \mathcal{I}[n])$

is an isomorphism for all $n \in \mathbf{Z}$. (Hom's taken in suitable derived categories.) By the adjointness of $f^{-1}$ and $Rf_*$, the adjointness of $Lg_!$ and $g^{-1}$, and their “primed” versions this follows from the isomorphism $g^{-1} Rf_*\mathcal{I} \to Rf'_* (g')^{-1}\mathcal{I}$ proved above.

In the case of a bounded complex $\mathcal{G}^\bullet$ (of modules or abelian groups) on $\mathcal{Y}_{lisse,{\acute{e}tale}}$ (resp. $\mathcal{Y}_{fppf}$) the canonical map

104.3.2.1
\begin{equation} \label{stacks-perfect-equation-to-show} L(g')_!(f')^{-1}\mathcal{G}^\bullet \to f^{-1}Lg_!\mathcal{G}^\bullet \end{equation}

is an isomorphism as follows from the case of a sheaf by the usual arguments involving truncations and the fact that the functors $L(g')_!(f')^{-1}$ and $f^{-1}Lg_!$ are exact functors of triangulated categories.

Suppose that $\mathcal{G}^\bullet$ is a bounded above complex (of modules or abelian groups) on $\mathcal{Y}_{lisse,{\acute{e}tale}}$ (resp. $\mathcal{Y}_{fppf}$). The canonical map (104.3.2.1) is an isomorphism because we can use the stupid truncations $\sigma _{\geq -n}$ (see Homology, Section 12.15) to write $\mathcal{G}^\bullet$ as a colimit $\mathcal{G}^\bullet = \mathop{\mathrm{colim}}\nolimits \mathcal{G}_ n^\bullet$ of bounded complexes. This gives a distinguished triangle

$\bigoplus \nolimits _{n \geq 1} \mathcal{G}_ n^\bullet \to \bigoplus \nolimits _{n \geq 1} \mathcal{G}_ n^\bullet \to \mathcal{G}^\bullet \to \ldots$

and each of the functors $L(g')_!$, $(f')^{-1}$, $f^{-1}$, $Lg_!$ commutes with direct sums (of complexes).

If $\mathcal{G}^\bullet$ is an arbitrary complex (of modules or abelian groups) on $\mathcal{Y}_{lisse,{\acute{e}tale}}$ (resp. $\mathcal{Y}_{fppf}$) then we use the canonical truncations $\tau _{\leq n}$ (see Homology, Section 12.15) to write $\mathcal{G}^\bullet$ as a colimit of bounded above complexes and we repeat the argument of the paragraph above.

Finally, by the adjointness of $f^{-1}$ and $Rf_*$, the adjointness of $Lg_!$ and $g^{-1}$, and their “primed” versions we conclude that the first identity of the lemma follows from the second in full generality. $\square$

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